question_answer

    A ball is hit at\[45{}^\circ \]to the horizontal with a kinetic energy\[{{E}_{k}}\]. The kinetic energy at the highest point is

A)  \[{{E}_{k}}\]                                     

B)  \[\frac{{{E}_{k}}}{2}\]

C)  \[\frac{{{E}_{k}}}{\sqrt{2}}\]                                     

D)  zero

Correct Answer: B

Solution :

                Key Idea: At the highest point vertical component of velocity is zero. Kinetic energy is possessed due to velocity. If m is mass of ball, then kinetic energy is                 \[{{E}_{k}}=\frac{1}{2}m{{u}^{2}}\] At the highest point or path only horizontal component of velocity exists, hence kinetic energy is,                 \[E{{}_{k}}=\frac{1}{2}m{{(u\cos 45{}^\circ )}^{2}}=\frac{1}{2}m{{\left( \frac{u}{\sqrt{2}} \right)}^{2}}\] \[\therefore \]  \[E{{}_{k}}=\frac{1}{2}\left( \frac{1}{2}m{{u}^{2}} \right)=\frac{{{E}_{k}}}{2}\]